Minimize $S=\dfrac{x}{y^2}+\dfrac{y}{z^2}+\dfrac{z}{x^2}$ when $x+y+z=xyz$.

Question

Let $x,y,z$ are positive real numbers that satisfy $x+y+z=xyz$. Find the minimum value of $$S=\dfrac{x}{y^2}+\dfrac{y}{z^2}+\dfrac{z}{x^2}$$

---

no minima found or $S_{Min}=1,73205...$

Answer 1

If $x=y=z=\sqrt3$ we get a value $\sqrt3$.

We'll prove that it's a minimal value of $S$.

Id est, we need to prove that $$\sum_{cyc}\frac{x}{y^2}\geq\sqrt{\frac{3(x+y+z)}{xyz}}.$$ Indeed, $(x,y,z)$ and $\left(\frac{1}{x^2},\frac{1}{y^2},\frac{1}{z^2}\right)$ are opposite ordered.

Thus, by Rearrangement $$\sum_{cyc}\frac{x}{y^2}=x\cdot\frac{1}{y^2}+y\cdot\frac{1}{z^2}+z\cdot\frac{1}{x^2}\geq x\cdot\frac{1}{x^2}+y\cdot\frac{1}{y^2}+z\cdot\frac{1}{z^2}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$ Thus, it remains to prove that $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq\sqrt{\frac{3(x+y+z)}{xyz}}$$ or $$xy+xz+yz\geq\sqrt{3xyz(x+y+z)},$$ which after squaring of the both sides gives $$\sum\limits_{cyc}z^2(x-y)^2\geq0.$$

Done!